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$\frac{1}{1 + a^{(n - m)}} + \frac{1}{1 + a^{(m - n)}} = ?$ $\frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} = ?$

10
2¹/₂
31
4a^{m + n}

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{1}{1 + a^{(n - m)}} + \frac{1}{1 + a^{(m - n)}} \\ = \frac{1}{(1 + \frac{a^{n}}{a^{m}})} + \frac{1}{(1 + \frac{a^{m}}{a^{n}})} \\ = \frac{a^{m}}{(a^{m} + a^{n})} + \frac{a^{n}}{(a^{m} + a^{n})} \\ = \frac{(a^{m} + a^{n})}{(a^{m} + a^{n})} \\ = 1 \end{aligned}

$\eqalign{ & \frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} \cr & = \frac{1}{{\left( {1 + \frac{{{a^n}}}{{{a^m}}}} \right)}} + \frac{1}{{\left( {1 + \frac{{{a^m}}}{{{a^n}}}} \right)}} \cr & = \frac{{{a^m}}}{{\left( {{a^m} + {a^n}} \right)}} + \frac{{{a^n}}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = \frac{{\left( {{a^m} + {a^n}} \right)}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = 1 \cr}$

$\frac{{(243)}^{n / 5} \times 3^{2 n + 1}}{9^{n} \times 3^{n - 1}} = ?$ $\frac{{{{\left( {243} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{9^n} \times {3^{n - 1}}}} = ?$

11
22
39
43ⁿ

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given Expression \\ = \frac{{(243)}^{n / 5} \times 3^{2 n + 1}}{9^{n} \times 3^{n - 1}} \\ = \frac{{(3^{5})}^{n / 5} \times 3^{2 n + 1}}{{(3^{2})}^{n} \times 3^{n - 1}} \\ = \frac{(3^{5 \times (n / 5)} \times 3^{2 n + 1})}{(3^{2 n} \times 3^{n - 1})} \\ = \frac{3^{n} \times 3^{2 n + 1}}{3^{2 n} \times 3^{n - 1}} \\ = \frac{3^{(n + 2 n + 1)}}{3^{(2 n + n - 1)}} \\ = \frac{3^{3 n + 1}}{3^{3 n - 1}} \\ = 3^{(3 n + 1 - 3 n + 1)} \\ = 3^{2} \\ = 9 \end{aligned}

$\eqalign{ & {\text{Given}}\,{\text{Expression}} \cr & = \frac{{{{\left( {243} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{9^n} \times {3^{n - 1}}}} \cr & = \frac{{{{\left( {{3^5}} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{{\left( {{3^2}} \right)}^n} \times {3^{n - 1}}}} \cr & = \frac{{\left( {{3^{5 \times \left( {n/5} \right)}} \times {3^{2n + 1}}} \right)}}{{\left( {{3^{2n}} \times {3^{n - 1}}} \right)}} \cr & = \frac{{{3^n} \times {3^{2n + 1}}}}{{{3^{2n}} \times {3^{n - 1}}}} \cr & = \frac{{{3^{\left( {n + 2n + 1} \right)}}}}{{{3^{\left( {2n + n - 1} \right)}}}} \cr & = \frac{{{3^{3n + 1}}}}{{{3^{3n - 1}}}} \cr & = {3^{\left( {3n + 1 - 3n + 1} \right)}} \cr & = {3^2} \cr & = 9 \cr}$

03. (0.04)^-1.5 = ?

125
2125
3250
4625

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} {(0.04)}^{- 1.5} = {(\frac{4}{100})}^{- 1.5} \\ = {(\frac{1}{25})}^{- (3 / 2)} \\ = {(25)}^{(3 / 2)} \\ = {(5^{2})}^{(3 / 2)} \\ = {(5)}^{2 \times (3 / 2)} \\ = 5^{3} \\ = 125 \end{aligned}

$\eqalign{ & {\left( {0.04} \right)^{ - 1.5}} = {\left( {\frac{4}{{100}}} \right)^{ - 1.5}} \cr & = {\left( {\frac{1}{{25}}} \right)^{ - \left( {3/2} \right)}} \cr & = {\left( {25} \right)^{\left( {3/2} \right)}} \cr & = {\left( {{5^2}} \right)^{\left( {3/2} \right)}} \cr & = {\left( 5 \right)^{2 \times \left( {3/2} \right)}} \cr & = {5^3} \cr & = 125 \cr}$

$\begin{aligned} \frac{1}{1 + x^{(b - a)} + x^{(c - a)}} + \\ \frac{1}{1 + x^{(a - b)} + x^{(c - b)}} + \\ \frac{1}{1 + x^{(b - c)} + x^{(a - c)}} = ? \end{aligned}$ $\eqalign{ & \frac{1}{{1 + {x^{\left( {b - a} \right)}} + {x^{\left( {c - a} \right)}}}} + \cr & \frac{1}{{1 + {x^{\left( {a - b} \right)}} + {x^{\left( {c - b} \right)}}}} + \cr & \frac{1}{{1 + {x^{\left( {b - c} \right)}} + {x^{\left( {a - c} \right)}}}} = ? \cr}$

10
21
3x^{a - b - c}
4None of these

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given exp . = \\ \frac{1}{(1 + \frac{x^{b}}{x^{a}} + \frac{x^{c}}{x^{a}})} + \frac{1}{(1 + \frac{x^{a}}{x^{b}} + \frac{x^{c}}{x^{b}})} + \frac{1}{(1 + \frac{x^{b}}{x^{c}} + \frac{x^{a}}{x^{c}})} \\ = \frac{x^{a}}{(x^{a} + x^{b} + x^{c})} + \frac{x^{b}}{(x^{a} + x^{b} + x^{c})} + \frac{x^{c}}{(x^{a} + x^{b} + x^{c})} \\ = \frac{(x^{a} + x^{b} + x^{c})}{(x^{a} + x^{b} + x^{c})} \\ = 1 \end{aligned}

$\eqalign{ & {\text{Given}}\,{\text{exp}}{\text{.}} = \cr & \frac{1}{{\left( {1 + \frac{{{x^b}}}{{{x^a}}} + \frac{{{x^c}}}{{{x^a}}}} \right)}} + \frac{1}{{\left( {1 + \frac{{{x^a}}}{{{x^b}}} + \frac{{{x^c}}}{{{x^b}}}} \right)}} + \frac{1}{{\left( {1 + \frac{{{x^b}}}{{{x^c}}} + \frac{{{x^a}}}{{{x^c}}}} \right)}} \cr & = \frac{{{x^a}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + \frac{{{x^b}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + \frac{{{x^c}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} \cr & = \frac{{\left( {{x^a} + {x^b} + {x^c}} \right)}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} \cr & = 1 \cr}$

05. The value of [(10)¹⁵⁰ ÷ (10)¹⁴⁶]

11000
210000
3100000
410⁶

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} {(10)}^{150} \div {(10)}^{146} = \frac{10^{150}}{10^{146}} \\ = 10^{150 - 146} \\ = 10^{4} \\ = 10000 \end{aligned}

$\eqalign{ & {\left( {10} \right)^{150}} \div {\left( {10} \right)^{146}} = \frac{{{{10}^{150}}}}{{{{10}^{146}}}} \cr & = {10^{150 - 146}} \cr & = {10^4} \cr & = 10000 \cr}$

06. (256)^0.16 x (256)^0.09 = ?

14
216
364
4256.25

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Answer:- 1

Explanation:-

Solution:

(256)^0.16 x (256)^0.09 = (256)^{(0.16 + 0.09)}

\begin{aligned} = {(256)}^{0.25} \\ = {(256)}^{\frac{25}{100}} \\ = {(256)}^{\frac{1}{4}} \\ = {(4^{4})}^{\frac{1}{4}} \\ = 4^{1} \\ = 4 \end{aligned}

$\eqalign{ & = {\left( {256} \right)^{0.25}} \cr & = {\left( {256} \right)^{\frac{{25}}{{100}}}} \cr & = {\left( {256} \right)^{\frac{1}{4}}} \cr & = {\left( {{4^4}} \right)^{\frac{1}{4}}} \cr & = {4^1} \cr & = 4 \cr}$

07. If 3^{(x - y)} = 27 and 3^{(x + y)} = 243, then x is equal to:

10
22
34
46

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Answer:- 1

Explanation:-

Solution:

3^{x - y} = 27 = 3³ ⇔ x - y = 3 ....(i)
3^{x + y} = 243 = 3⁵ ⇔ x + y = 5 ....(ii)
On solving (i) and (ii), we get x = 4.

08. Given that 10^0.48 = x, 10^0.70 = y and x^z = y², then the value of z is close to:

11.45
21.88
32.9
43.7

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} x^{2} = y^{2} \Leftrightarrow 10^{(0.48 z)} = 10^{2 \times 0.70} = 10^{1.40} \\ \Rightarrow 0.48 z = 1.40 \\ \Rightarrow z = \frac{140}{48} = \frac{35}{12} = 2.9 (approx) \end{aligned}

$\eqalign{ & {x^2} = {y^2} \Leftrightarrow {10^{\left( {0.48z} \right)}} = {10^{2 \times 0.70}} = {10^{1.40}} \cr & \Rightarrow 0.48z = 1.40 \cr & \Rightarrow z = \frac{{140}}{{48}} = \frac{{35}}{{12}} = 2.9({\text{approx}}) \cr}$

$\begin{aligned} If {(\frac{a}{b})}^{x - 1} = {(\frac{b}{a})}^{x - 3}, \\ then the value of x is \end{aligned}$ $\eqalign{ & {\text{If}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}}, \cr & {\text{then}}\,{\text{the}}\,{\text{value}}\,{\text{of}}\,x\,{\text{is}} \cr}$

1¹/₂
21
32
4⁷/₂

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given {(\frac{a}{b})}^{x - 1} = {(\frac{b}{a})}^{x - 3} \\ \Rightarrow {(\frac{a}{b})}^{x - 1} = {(\frac{a}{b})}^{- (x - 3)} = {(\frac{a}{b})}^{(3 - x)} \\ \Rightarrow x - 1 = 3 - x \\ \Rightarrow 2 x = 4 \\ \Rightarrow x = 2 \end{aligned}

$\eqalign{ & {\text{Given}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{a}{b}} \right)^{ - \left( {x - 3} \right)}} = {\left( {\frac{a}{b}} \right)^{\left( {3 - x} \right)}} \cr & \Rightarrow x - 1 = 3 - x \cr & \Rightarrow 2x = 4 \cr & \Rightarrow x = 2 \cr}$

10. (17)^3.5 x (17)^? = 17⁸

12.29
22.75
34.25
44.5

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Answer:- 1

Explanation:-

Solution:

Let (17)^3.5 x (17)^x = 17⁸.
Then, (17)^{3.5 + x} = 17⁸.
∴ 3.5 + x = 8
⇒ x = (8 - 3.5)
⇒ x = 4.5

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Surds and Indices - 05

$\frac{1}{1 + a^{(n - m)}} + \frac{1}{1 + a^{(m - n)}} = ?$ $\frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} = ?$

$\frac{{(243)}^{n / 5} \times 3^{2 n + 1}}{9^{n} \times 3^{n - 1}} = ?$ $\frac{{{{\left( {243} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{9^n} \times {3^{n - 1}}}} = ?$

03. (0.04)^-1.5 = ?

05. The value of [(10)¹⁵⁰ ÷ (10)¹⁴⁶]

06. (256)^0.16 x (256)^0.09 = ?

07. If 3^{(x - y)} = 27 and 3^{(x + y)} = 243, then x is equal to:

08. Given that 10^0.48 = x, 10^0.70 = y and x^z = y², then the value of z is close to:

10. (17)^3.5 x (17)^? = 17⁸

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NCERT
6 to 12

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Surds and Indices - 05

01. 11+a(n−m)+11+a(m−n)=?11+a(n−m)+11+a(m−n)=?\frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} = ?

02. (243)n/5×32n+19n×3n−1=?(243)n/5×32n+19n×3n−1=?\frac{{{{\left( {243} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{9^n} \times {3^{n - 1}}}} = ?

03. (0.04)-1.5 = ?

05. The value of [(10)150 ÷ (10)146]

06. (256)0.16 x (256)0.09 = ?

07. If 3(x - y) = 27 and 3(x + y) = 243, then x is equal to:

08. Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:

09. If(ab)x−1=(ba)x−3,thenthevalueofxisIf(ab)x−1=(ba)x−3,thenthevalueofxis\eqalign{ & {\text{If}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}}, \cr & {\text{then}}\,{\text{the}}\,{\text{value}}\,{\text{of}}\,x\,{\text{is}} \cr}

10. (17)3.5 x (17)? = 178

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NCERT 6 to 12

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Practice Batch

$\frac{1}{1 + a^{(n - m)}} + \frac{1}{1 + a^{(m - n)}} = ?$ $\frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} = ?$

$\frac{{(243)}^{n / 5} \times 3^{2 n + 1}}{9^{n} \times 3^{n - 1}} = ?$ $\frac{{{{\left( {243} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{9^n} \times {3^{n - 1}}}} = ?$

03. (0.04)^-1.5 = ?

05. The value of [(10)¹⁵⁰ ÷ (10)¹⁴⁶]

06. (256)^0.16 x (256)^0.09 = ?

07. If 3^{(x - y)} = 27 and 3^{(x + y)} = 243, then x is equal to:

08. Given that 10^0.48 = x, 10^0.70 = y and x^z = y², then the value of z is close to:

10. (17)^3.5 x (17)^? = 17⁸

NCERT
6 to 12